Electrical Revolution: Advantages of High Transmission Line Voltage

Advantages of high voltage transmission line

( a ) Volume of Conductor Material

Let us consider that the electrical power transmitted through transmission line is

P = √ 3 V_LI_L Cos Φ

I_L = P / √ 3 V_L Cos Φ…. ( 1 )

Where

V_L = Line voltage

I_L = Line current

P = Power transmitted

Cos Φ = Power factor of the load

Resistance per conductor R = ρL / a……… ( 2 )

Total power loss W = 3I²R

= 3 ( P / √ 3 V_L Cos Φ )² ( ρL / a )

= 3 ( P² / 3 V_L² Cos ²Φ ) ( ρL / a )

W = ( P² ρL / a V_L² Cos ²Φ ) …….. ( 3 )

Cross section area a = ( P² ρL / W V_L² Cos ²Φ )

Volume of conductor = Area of conductor × Length of conductor

= aL

= ( P² ρL² / W V_L² Cos ²Φ )……..( 4 )

Volume of conductor α ( 1 / V_L² Cos ²Φ )

The volume of conductor is inversely square of supply voltage and inversely proportional to square of power factor if power transmitted, length of conductor, transmission line losses and resistivity of material is remains constant.

( b ) Decrease in voltage drop

Voltage drop = IR

= I ( ρL / a )

= I ( ρLJ / I ) ( Current density J = I / a )

= ρLJ

% Voltage drop = [ ρLJ / V ] × 100%

We can say that If the voltage increases, percentage voltage drop decreases.

( c ) Transmission line efficiency

Input power = Pi

Output power = P

Input power = Output Power + Losses

= P + ( P² ρL / a V_L² Cos ²Φ )

= P + ( P² ρLJ / I V_L² Cos ²Φ ) ( current density J = I / a )

As P = √ 3 V_L I_L Cos Φ

= P + { P² ρLJ ( √ 3 V_L Cos Φ ) / P V_L² Cos ²Φ }

= P + { P ρLJ ( √ 3 ) / V_L Cos Φ }

= P + { ( √ 3 ) P ρLJ / V_L Cos Φ }

= P { 1 + [ √ 3 ρLJ / V_L Cos Φ ] }

Transmission line efficiency = [ Output power / Input power ] × 100%

= P / P { 1 + [ √ 3 ρLJ / V_L Cos Φ ] }× 100%

= { 1 – [ √ 3 ρLJ / V_L Cos Φ ] }× 100%

If current density, resistivity and length of line is constant the efficiency of the transmission line is inversely proportional to line voltage and power factor of the load.

Conclusion

If the transmission line voltage increases,